Question

make a sketch for each problem. label the lengths that are known. then use the pythagorean theorem to solve the problem.
a baseball diamond is a square 90 feet on each side. how long is a throw from home plate to second base (to the nearest foot)?
the bottom of a 25 - foot ladder is placed 7 feet from a wall. how far up the wall will the ladder reach?
a path leads across the park from one corner to the opposite corner. the park is 150 meters wide and 200 meters long. how far would you walk if you took the path instead of walking around? how many meters would you save?
sam wants to brace a bookcase by nailing a strip of wood from the lower left corner to the upper right corner. the bookcase is 1 meter wide and 2 meters high. how long should the brace be (to the nearest hundredth of a meter)?

Explanation:

1. Baseball diamond problem

Step1: Identify the right - triangle

A baseball diamond is a square with side length \(a = 90\) feet. The throw from home plate to second base is the diagonal of the square. The square can be divided into two right - triangles, where the two legs of the right - triangle have length \(a = 90\) feet.

Step2: Apply the Pythagorean Theorem

The Pythagorean Theorem is \(c^{2}=a^{2}+b^{2}\). In a square, \(a = b=90\) feet. So \(c^{2}=90^{2}+90^{2}=8100 + 8100=16200\).

Step3: Solve for \(c\)

\(c=\sqrt{16200}\approx127.28\approx127\) feet.

2. Ladder problem

Step1: Identify the right - triangle

The ladder, the wall, and the ground form a right - triangle. The length of the ladder is the hypotenuse \(c = 25\) feet and the distance from the bottom of the ladder to the wall is \(a = 7\) feet. Let the height up the wall that the ladder reaches be \(b\).

Step2: Apply the Pythagorean Theorem

According to the Pythagorean Theorem \(c^{2}=a^{2}+b^{2}\), so \(b^{2}=c^{2}-a^{2}\).

Step3: Substitute values

Substitute \(c = 25\) and \(a = 7\) into the formula: \(b^{2}=25^{2}-7^{2}=625 - 49 = 576\).

Step4: Solve for \(b\)

\(b=\sqrt{576}=24\) feet.

3. Park path problem

Step1: Identify the right - triangle

The park is a rectangle with length \(a = 200\) meters and width \(b = 150\) meters. The path across the park is the diagonal of the rectangle.

Step2: Apply the Pythagorean Theorem

The length of the diagonal \(c\) of the rectangle is given by \(c^{2}=a^{2}+b^{2}\), so \(c^{2}=200^{2}+150^{2}=40000+22500 = 62500\).

Step3: Solve for \(c\)

\(c=\sqrt{62500}=250\) meters. The distance walking around is \(200 + 150=350\) meters. The distance saved is \(350 - 250 = 100\) meters.

4. Bookcase bracing problem

Answer:

127 feet